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If-cos-1-x-cos-1-y-cos-1-z-cos-1-u-2pi-then-x-1999-y-2000-z-2001-u-2002-




Question Number 103155 by 9027201563 last updated on 13/Jul/20
If cos^(−1) x+cos^(−1) y+cos^(−1) z+cos^(−1) u=2π,  then  x^(1999) +y^(2000) +z^(2001) +u^(2002) =
$$\mathrm{If}\:\mathrm{cos}^{−\mathrm{1}} {x}+\mathrm{cos}^{−\mathrm{1}} {y}+\mathrm{cos}^{−\mathrm{1}} {z}+\mathrm{cos}^{−\mathrm{1}} {u}=\mathrm{2}\pi, \\ $$$$\mathrm{then}\:\:{x}^{\mathrm{1999}} +{y}^{\mathrm{2000}} +{z}^{\mathrm{2001}} +{u}^{\mathrm{2002}} = \\ $$
Commented by 9027201563 last updated on 13/Jul/20
solution needed urgently
$$\mathrm{solution}\:\mathrm{needed}\:\mathrm{urgently} \\ $$
Commented by prakash jain last updated on 13/Jul/20
I dont think you can get a unique  answer  assuming  cos^(−1) x=cos^(−1) y=π  cos^(−1) z=cos^(−1) u=0  x=−1,y=−1,z=1,u=1  sum=2  cos^(−1) x=cos^(−1) y=cos^(−1) z=cos^(−1) u=(π/2)  x=y=z=u=0  sum=0
$$\mathrm{I}\:\mathrm{dont}\:\mathrm{think}\:\mathrm{you}\:\mathrm{can}\:\mathrm{get}\:\mathrm{a}\:\mathrm{unique} \\ $$$$\mathrm{answer} \\ $$$$\mathrm{assuming} \\ $$$$\mathrm{cos}^{−\mathrm{1}} {x}=\mathrm{cos}^{−\mathrm{1}} {y}=\pi \\ $$$$\mathrm{cos}^{−\mathrm{1}} {z}=\mathrm{cos}^{−\mathrm{1}} {u}=\mathrm{0} \\ $$$${x}=−\mathrm{1},{y}=−\mathrm{1},{z}=\mathrm{1},{u}=\mathrm{1} \\ $$$${sum}=\mathrm{2} \\ $$$$\mathrm{cos}^{−\mathrm{1}} {x}=\mathrm{cos}^{−\mathrm{1}} {y}=\mathrm{cos}^{−\mathrm{1}} {z}=\mathrm{cos}^{−\mathrm{1}} {u}=\frac{\pi}{\mathrm{2}} \\ $$$${x}={y}={z}={u}=\mathrm{0} \\ $$$${sum}=\mathrm{0} \\ $$

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